A function $f: [a,b] \rightarrow \mathbb{R}$ is Lipschitz continuous if and only if $\forall \epsilon >0, \exists \delta>0$ such that for every finite collection of closed intervals in $[a,b]$, $\{[a_k,b_k]\}_{k=1}^N$ (not necessarily disjoint) with
$\sum_{k=1}^N b_k-a_k < \delta$ then $\sum_{k=1}^N |f(b_k)-f(a_k)|<\epsilon$
The easy part is to prove that a Lipschitz function satisfy the property. But I'm having trouble with the other part.
I tried to mimic the proof that $\sqrt{x}$ does not satisfy the property with not necessarily disjoint intervals but unfortunately I got stuck. I saw this exercise on chapter 5 of Bruckner, real analysis.
I would really appreciate any hints or suggestions.
Suppose $f$ is not Lipschitz continuous. Let $\epsilon>0$ be given. I will construct intervals with $\sum |a_k-b_k|<\epsilon$ but $$\sum|f(a_k)-f(b_k)|\ge 1/2\tag1$$
Since $f$ is not Lipschitz, there exist points $s<t$ such that $$|f(s)-f(t)|>\epsilon^{-1}|s-t|\tag2$$
Case 1: $|s-t|<\epsilon$.
Use the interval $[s,t]$ as $[a_k,b_k]$, repeating it $N$ times where $N$ is chosen so that $$\frac{\epsilon}{2} \le N|s-t| < \epsilon\tag3$$ (Such $N$ always exists, why?) Observe that (2) and (3) imply (1).
Case 2: $|s-t| \ge \epsilon$.
Divide the interval $[s,t]$ into $N$ equal subintervals $[s_k,t_k]$; choose the number of pieces $N$ so that $$\frac{\epsilon}{2}\le \frac{|s-t|}{N} < \epsilon\tag4$$ (This is always possible, why?) The triangle inequality implies that at least one piece $[s_k,t_k]$ satisfies $$|f(s_k)-f(t_k)|>\epsilon^{-1}|s_k-t_k|\tag5$$ Use the single interval $[s_k,t_k]$ to achieve (1), using (4) and (5).